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HW Sol Due 3-10-2009

HW Sol Due 3-10-2009 - HW Due 3-10—2009 Problem 5.3 Find...

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Unformatted text preview: HW Due 3-10—2009 Problem 5.3: Find the system performance measures of CT, WIP and throughput for a serial system consisting of 3 single capacity workstations. The arrival rate of jobs is 3/hr, with the inter-arrival times being exponentially distributed. The workstation processing time data is: Workstation i 2 ' E[T.] | C [T.] I l HW Sol Due 3-10—2009 Hw 5.3 File: HW5p3.mle WS 1:1am1: 3 C2al: 1 W81: U1: 0.75 Cqu: 1.88 CTSl: 2.13 WIPsl: 6.38 C2d1: 2.69 WS 2:1am2: 3 C2a2: 2.69 WSZ: U2: 0.87 CTq2: 5.52 CTSZ: 5.81 WIPSZ: 17.43 C2d2: 2.92 WS 2: 131113: 3 C283: 2.92 W83: U3: 0.90 CTq3: 6.65 CTs3: 6.95 WIPs3: 20.84 C2d3: 2.18 System CTS: 14.88 ths: 3 WIPS: 44.64 Problem 3 Due to the serial nature of the system: Al=lz=7L3=3 Workstation 1: _ llE[T1] _ 3(0.25) c; I CI; (1) = (flIO—ZS—JOZS + 0.25 = 2.125 2 0.25 W113. (1) = 71107; (1) = 3(2.125) = 6.375 Cj(1)=(1— aflcg (1) + 773—03 (1) = (1 — 0.752 )1 4 0.752(4) = 2.6875 “1 = 0.75 Workstation 2: _ AZEU'Z] _ 3(0.29) “2 = 0.87 :32 1 07;. (2) =' (MIQEJO29 + 0.29 = 5.809 2 0.13 WIPE (2) = 71267; (2) = 3(5809) = 17.427 (33(2) = (1 —u§)c§(2)+u§Cf(2) = (1 — 0.872)2.6875+ 0.872(3) = 2.924 Workstation 3: 7L3E[T3 3(0.30) _ ] u3———— c3 1 C]; (3) = [2922442134193030 + 0.30 = 6.9474 W113 (3) = 71367;. (3) = 3(6.9474) = 20.842 03 (3) = (1 — u§)C§(3)+ u§C§ (3) = (1 — 0.902)2.924+ 0.902(2) = 2.1756 = 0.90 W31)»: = W11”: (0+ Wffl(2)+ W136) = 44.645 Thays : A] = 3 CT _ Ways _ 44.645 sys Th”: = 14.882 ...
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